p34_053 - 53 Let θ1 = 45◦ be the angle of incidence at...

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Unformatted text preview: 53. Let θ1 = 45◦ be the angle of incidence at the first surface and θ2 be the angle of refraction there. Let θ3 be the angle of incidence at the second surface. The condition for total internal reflection at the second surface is n sin θ3 ≥ 1. We want to find the smallest value of the index of refraction n for which this inequality holds. The law of refraction, applied to the first surface, yields n sin θ2 = sin θ1 . Consideration of the triangle formed by the surface of the slab and the ray in the slab tells us that θ3 = 90◦ − θ2 . Thus, the condition for total internal reflection becomes 1 ≤ n sin(90◦ − θ2 ) = n cos θ2 . Squaring this equation and using sin2 θ2 + cos2 θ2 = 1, we obtain 1 ≤ n2 (1 − sin2 θ2 ). Substituting sin θ2 = (1/n) sin θ1 now leads to sin2 θ1 1 ≤ n2 1 − = n2 − sin2 θ1 . n2 The largest value of n for which this equation is true is the value for which 1 = n2 − sin2 θ1 . We solve for n: n = 1 + sin2 θ1 = 1 + sin2 45◦ = 1.22 . ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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